CALC Debye’s T 3 ?Law. At very low temperatures the molar heat capacity of rock salt varies with temperature according to Debye’s T3 law: where k = 1940 J/mol ? K and ? = 281 K. (a) How much heat is required to raise the temperature of 1.50 mol of rock salt from 10.0 K to 40.0 K? (?Hint: Use Eq. (17.18) in the form dQ = nCdT and integrate.) (b) What is the average molar heat capacity in this range? (c) What is the true molar heat capacity at 40.0 K?
Solution 95P Step 1 of 6: (a) How much heat is required to raise the temperature of 1.50 mol of rock salt from 10.0 K to 40.0 K (H int: Use Eq. (17.18) in the form dQ = nCdT and integrate.) Given data, Initial temperature,T =i10k Final temperature, T = 40k f Number of moles, n = 1.5 mol k= 1940 J/mol. K Debye temperature, = 281 k Molar specific heat is the energy required to raise the temperature of one mole of substance by one degree. That is given by 3 Molar heat capacity, c = k T3 Step 2 of 6: The equation that relates change in temperature and heat gain or loss, Q= n c T Where Q is the heat gain or loss, n is number of moles , c is molar specific heat and T is change in temperature. For differential Heat gain or loss is given by, dQ= n c dT Using c = k T 3 T3 dQ= n (k 3 ) dT Since specific heat is the function of temperature, to calculate the heat required ; we need to integrate above equation over temperature change, Integrating both sides, T f T f T3 dQ = n k 3 dT Ti Ti Step 3 of 6: Taking constant outside, Tf n k 3 Q = 3 T dT T i On integrating, Q = n k[ ]4 Tf 3 4 T i T T 4 Q = n k[ f i ] 3 4 Q = n k [T 4 T ]4 4 f i